18-Env-B5 Industrial & Hazardous Waste Management · December 2019
Question 3 of 10: Half-Life from a First-Order Degradation Rate Constant Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10
Notes on this paper
Reference texts: LaGrega, Buckingham & Evans, Hazardous Waste Management , 2nd ed.; Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment , 2nd ed.; Davis & Cornwell, Introduction to Environmental Engineering , 6th ed.; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery , 5th ed.; Cooper & Alley, Air Pollution Control: A Design Approach ; ACGIH, Industrial Ventilation: A Manual of Recommended Practice ; Ontario Environmental Protection Act , R.S.O. 1990, c. E.19 and O. Reg. 347 (Waste Management – General); U.S. EPA SW-846 Method 1311 (Toxicity Characteristic Leaching Procedure).
Question 3: Half-Life from a First-Order Degradation Rate Constant (10 marks)
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
Given. First-order degradation rate constant $k = 0.02\ \text{day}^{-1}$.
Find. The half-life $t_{1/2}$ of the compound under first-order decay.
Approach. Integrate the first-order decay law $C = C_0 e^{-kt}$ and solve for the time at which $C = C_0/2$.
Set up the first-order decay relation. For first-order kinetics, $$C = C_0 e^{-kt}$$ At the half-life, $C = C_0/2$, so $$\frac{1}{2} = e^{-k\,t_{1/2}}$$Solve for the half-life. Taking the natural log of both sides and rearranging gives the standard first-order half-life relation, $$t_{1/2} = \frac{\ln 2}{k}$$ Substituting $k = 0.02\ \text{day}^{-1}$: $$t_{1/2} = \frac{0.6931}{0.02\ \text{day}^{-1}} = \boxed{34.7\ \text{days}}$$
Final results
Quantity Value Rate constant, $k$ $0.02\ \text{day}^{-1}$ Half-life, $t_{1/2}$ 34.7 days ($\approx$ 4.95 weeks)
Topic: First-order degradation kinetics and half-life
Key relations: $C = C_0 e^{-kt}$; $t_{1/2} = \ln 2 / k = 0.693/k$, independent of the starting concentration $C_0$
Why this works: a first-order process removes a constant FRACTION of whatever mass remains per unit time, so the time to halve the concentration is the same regardless of how much is present — that is exactly what makes $t_{1/2}$ a single, concentration-independent number characterizing the compound's persistence in a given treatability test.
Common pitfall: using $t_{1/2} = 1/k$ (mistaking the mean residence time / time constant for the half-life) instead of the correct $\ln 2/k$ factor, or forgetting to keep the time units ($k$ in day$^{-1}$ gives $t_{1/2}$ directly in days).
Source: Davis & Cornwell, Introduction to Environmental Engineering , 6th ed.
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